Why can one calculate entropy change for thermal conduction?

A hot object in thermal contact with a cold one will finally reach a temperature in between. Why can the entropy change of each object be calculated as if the process was reversible? Is there a reversible process with the same final and initial state and what would that be?

Staff: Mentor

A hot object in thermal contact with a cold one will finally reach a temperature in between. Why can the entropy change of each object be calculated as if the process was reversible? Is there a reversible process with the same final and initial state and what would that be?

To determine the entropy change for an irreversible process, the first step is to TOTALLY FORGET ABOUT THE ACTUAL IRREVERSIBLE PROCESS THAT BROUGHT THE SYSTEM FROM ITS INITIAL STATE TO THE FINAL STATE, and focus only on the two end states.
Step 2: Devise a reversible path between the two end states. There are an infinite number of reversible paths, and they all give the same result for the entropy change, so any one will do. Choose one that is simple to apply step 3.
Step 3: Calculate the integral of dq/T for this reversible path.

This is what they mean when they say delta S is the integral of dqrev/T.

In the case of the hot object and the cold object, the final state has a temperature somewhere in-between the initial temperatures of the two. If I were doing Step 2, I would first separate the two objects, and then devise a reversible process to bring each of them to the final state individually. To do this, for each object, I would contact it with a continuous sequence of constant temperature reservoirs, each reservoir having a temperature slightly different from the previous one. Using these reservoirs, I would very gradually and reversibly bring each object to the final temperature that was attained in the irreversible process. In Step 3, I would then calculate the change in entropy of each object individually, and then add the two changes in entropy to get the overall change.